[2,3] 这道题给了我们一个数组,数组中的数字可能出现一次或两次,让我们找出所有出现两次的数字,由于之前做过一道类似的题目Find the Duplicate Number,所以不是完全无从下手。这类问题的一个重要条件就是1 ≤ a[i] ≤ n (n = size of array),不然很难在O(1)空间和O(n)时间内完成。首先来看一种正负...
If the last number in the sorted array is not the N (size of the array), the missing number can be simply set to N. This approach uses O(1) constant space. Set It is straightforward to use set (or preferably the unordered_set). The space complexity is O(N) and the time complexit...
How to find duplicate values in a JavaScript array - In this tutorial, we will discuss how we can find duplicate or repeating values in a JavaScript array using different methods or approaches to reach the solution to this problem. Below is the list of t
类似LeetCode 442. Find All Duplicates in an Array,由于元素是1~n,因此每个元素的值-1(映射到0~n-1)就可以直接当做下标。 classSolution {public:intfindDuplicate(vector<int>&nums) {for(inti=0;i<nums.size();++i){intindex=abs(nums[i])-1;if(nums[index]>0) nums[index]*=-1;elsereturnind...
There is only one duplicate number in the array, but it could be repeated more than once 非常好的题目,开始是用二分做的,比如取数组为{1,2,3,3,4,5},mid应该是(5+1)/2 = 3,那么,如果小于等于mid的数的个数如果超过了3,那么重复的数字一定出现在l,mid之间,否则出现在mid + 1,r之间。以该...
There is only one duplicate number in the array, but it could be repeated more than once. 思路 根据题意,题目给的是一个包含1到n+1的integer数组,其中不重复的数是(1-n)个,也就是说至少有2个数是重复。现在需要找出这个数来。 试想,给定一个数 m: ...
Note: You must not modify the array (assume the array is read only). You must use only constant, O(1) extra space. Your runtime complexity should be less than O(n2). There is only one duplicate number in the array, but it could be repeated more than once. ...
Find the Duplicate Number数组中重复的数 题意:数组中有1+n个数都是1到n之间,找出其中重复的。 解法一:二分搜索,先求出中点mid,然后遍历整个数组,统计所有小于等于mid的数的个数,如果个数小于等于mid,则说明重复值在[mid+1, n]之间,反之,重复值应在[1, mid-1]之间。 解法二:利用快慢指针,思路和带环...
- There is only one duplicate number in the array, but it could be repeated more than once. 代码如下: publicintfindDuplicate(int[] nums) {if(nums.length >1){intslow = nums[0];intfast = nums[nums[0]];//以下为求两个指针第一次相遇的点while(slow != fast){ ...
Since each value in a Set has to be unique, passing any duplicate item will be removed automatically:const numbers = [1, 2, 3, 2, 4, 5, 5, 6]; const unique = Array.from(new Set(numbers)); console.log(unique); // [ 1, 2, 3, 4, 5, 6 ] ...